HP71B Integral Questions

[Wes Loewer]

(02-08-2020 03:04 PM)Wes Loewer Wrote:  

(02-08-2020 10:59 AM)J-F Garnier Wrote:  Actually, the endpoints are not really left out, just it's useless to evaluate the function at the endpoints since their contribution to the sum is null due to the (1-u²) factor. So nothing is missing in the sum.

If the outside trapezoids don't contribute to the sum, isn't that equivalent to leaving them out? Wouldn't the approximate area of the outer intervals be missing from the sum?

I see now why 1/sqrt(1-x^2) has such a difficult time converging with this algorithm. Every time the algorithm goes to a smaller interval, the total width being considered keeps growing. It's like trying to hit a moving target.

For most functions, the cubic transformation will cause the endpoints to approach 0, so the added area has a shrinking height and width. But in this case, the endpoints approach a finite value (~1.73), so while the width is shrinking, the height is not.

This has all been very educational. :-)